Long-run proportion that harmonic light bulb is on

Question

Suppose a light bulb is turned on at time $t=0.$ It switches off at $t=1,$ on again at $t=1+{1 \over 2},$ off at $t=1+ {1 \over 2}+{1 \over 3},$ and so forth. As $t$ goes to $\infty,$ what proportion of the time is the light bulb on?

Answer 1

The whole series is the alternating harmonic series, which in the limit as n approaches infinity equals $ln 2$.

As both individual series diverge, the difference goes to a 0 fraction in the limit, and the ratio approaches 1.0. Therefore the light is on half the time.

Answer 2

Consider the first few terms of the series concerning the period of time the light is turned on and off:

$$\text{ON: }1, \frac 13, \frac 15,...\\ \text{OFF: } \frac 12,\frac 14, \frac 16,...$$

You can clearly see that the times you are looking for are:

$$\displaystyle\sum^{\infty}_{i=1} {\frac 1{2i-1}} $$

and

$$\displaystyle\sum^{\infty}_{i=1} {\frac 1{2i}} $$

Use harmonic series to solve

Answer 3

At any point in time, the light has been on longer than it has been off, as $1 > \frac{1}{2}$, $\frac{1}{3} > \frac{1}{4}$, etc. If we start looking at the light at $t = 1$, then at any point in time the light has been off longer than it has been on, as $\frac{1}{2} > \frac{1}{3}$, $\frac{1}{4} > \frac{1}{5}$, etc. So the proportion of the time that the light is on must be $\frac{1}{2}$.